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fiasKO [112]
3 years ago
13

Because sample means were being tested, the is used to calculate the z-score.

Mathematics
1 answer:
jasenka [17]3 years ago
5 0

Answer:

What?

Step-by-step explanation:

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Use the general solution to solve 5 – 6x = 8x + 17.
Slav-nsk [51]

Answer:

8x+6x=14x

5-17= -12

divide both sides by 2

7x= -6

divide by 7

you get x= -6 over 7

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Rectangle 1 has length x and width y. Rectangle 2 is made by multiplying each dimension of Rectangle 1 by a factor of k, where k
tresset_1 [31]

Answer:

a) Similar

b) Perimeter of rectangle 2 is k times the Perimeter of rectangle 1 (Proved Below)

c) Area of rectangle 2 is k^2 times the Area of rectangle 1 (Proved Below)

Step-by-step explanation:

Given:

Rectangle 1 has length = x

Rectangle 1 has width = y

Rectangle 2 has length = kx

Rectangle 2 has width = ky

(a) Are Rectangle 1 and Rectangle 2 similar? Why or why not?

Rectangle 1 and Rectangle 2 are similar because the angles of both rectangles are 90° and the sides of Rectangle 2 is k times the sides of Rectangle 1. So sides of both rectangles is equal to the ratio k.

(b) Write a paragraph proof to show that the perimeter of Rectangle 2 is k times the perimeter of Rectangle 1.

Perimeter of Rectangle = 2*(Length + Width)

Perimeter of Rectangle 1 = 2*(x+y) = 2x+2y

Perimeter of Rectangle 2 = 2*(kx+ky) = 2kx + 2ky

                                          = k(2x+2y)

                                          = k(Perimeter of Rectangle 1)

Hence proved that Perimeter of rectangle 2 is k times the perimeter of rectangle 1.

(c) Write a paragraph proof to show that the area of Rectangle 2 is k^2 times the area of Rectangle 1.

Area of Rectangle = Length * width

Area of Rectangle 1 = x * y

Area of Rectangle 2 = kx*ky

                                  = k^2 (xy)

                                  = k^2 (Area of rectangle 1)

Hence proved that area of rectangle 2 is k^2 times the area of rectangle 1.

4 0
3 years ago
You are asked to do a study of shelters for abused and battered women to determine the necessary capacity in your city to provid
ohaa [14]

Answer:

Using the normal probability distribution, with a capacity of 350, it is enough for all abused on 90.82% of nights.

274 shelters will be needed.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 250, \sigma = 75

If the city’s shelters have a capacity of 350, will that be enough places for abused women on 95% of all nights?

What is the percentile of 350?

This is the pvalue of Z when X = 350.

Z = \frac{X - \mu}{\sigma}

Z = \frac{250 - 150}{75}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082.

Using the normal probability distribution, with a capacity of 350, it is enough for all abused on 90.82% of nights.

If not, what number of shelter openings will be needed?

The 95th percentile, which is X when Z = 1.645. So

Z = \frac{X - \mu}{\sigma}

1.645 = \frac{X - 150}{75}

X - 150 = 1.645*75

X = 274

274 shelters will be needed.

5 0
3 years ago
What is the volume of a sphere with radius of 16 inches? Use 3.14 for π. 200.96 in3 17,148.59 in3 1071.79 in3 4287.15 in3
crimeas [40]

Answer:

17148.59 in^3

Step-by-step explanation:

This volume is calculated as follows:  V = (4/3)(3.14)(16 in)^3 = 17148.95 in^3

3 0
3 years ago
Read 2 more answers
Matt gave half of his books to the local library and kept the other half. His best friend gave him 3 more books. He now has 57 b
Pani-rosa [81]
He started with 108 books
5 0
3 years ago
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